YES
The TRS could be proven terminating. The proof took 365 ms.
The following DP Processors were used
Problem 1 was processed with processor DependencyGraph (17ms).
| – Problem 2 was processed with processor PolynomialOrderingProcessor (122ms).
| – Problem 3 was processed with processor SubtermCriterion (1ms).
Problem 1: DependencyGraph
Dependency Pair Problem
Dependency Pairs
| minus#(x, s(y)) | → | p#(minus(x, p(s(y)))) | | minus#(x, s(y)) | → | minus#(x, p(s(y))) |
| le#(s(x), s(y)) | → | le#(x, y) | | minus#(x, s(y)) | → | if#(le(x, s(y)), 0, p(minus(x, p(s(y))))) |
| minus#(x, s(y)) | → | p#(s(y)) | | minus#(x, s(y)) | → | le#(x, s(y)) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(x, s(y)) | → | if(le(x, s(y)), 0, p(minus(x, p(s(y))))) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, le, if, p, true, false
Strategy
The following SCCs where found
| minus#(x, s(y)) → minus#(x, p(s(y))) |
| le#(s(x), s(y)) → le#(x, y) |
Problem 2: PolynomialOrderingProcessor
Dependency Pair Problem
Dependency Pairs
| minus#(x, s(y)) | → | minus#(x, p(s(y))) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(x, s(y)) | → | if(le(x, s(y)), 0, p(minus(x, p(s(y))))) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, le, if, p, true, false
Strategy
Polynomial Interpretation
- 0: -2
- false: -2
- if(x,y,z): -2
- le(x,y): -2
- minus(x,y): -2
- minus#(x,y): y - 1
- p(x): x - 2
- s(x): 2x + 2
- true: -2
Improved Usable rules
The following dependency pairs are strictly oriented by an ordering on the given polynomial interpretation, thus they are removed:
| minus#(x, s(y)) | → | minus#(x, p(s(y))) |
Problem 3: SubtermCriterion
Dependency Pair Problem
Dependency Pairs
| le#(s(x), s(y)) | → | le#(x, y) |
Rewrite Rules
| p(0) | → | 0 | | p(s(x)) | → | x |
| le(0, y) | → | true | | le(s(x), 0) | → | false |
| le(s(x), s(y)) | → | le(x, y) | | minus(x, 0) | → | x |
| minus(x, s(y)) | → | if(le(x, s(y)), 0, p(minus(x, p(s(y))))) | | if(true, x, y) | → | x |
| if(false, x, y) | → | y |
Original Signature
Termination of terms over the following signature is verified: 0, minus, s, le, if, p, true, false
Strategy
Projection
The following projection was used:
Thus, the following dependency pairs are removed:
| le#(s(x), s(y)) | → | le#(x, y) |